Going off the example the following question (The Rankine-Hugoniot condition), how do we calculate the Rankine Hugoniot condition for nonscalar initial data? For $t \leq 1$ I understand how the author obtains the piecewise solution, but I am unclear how he obtains the bounds for each solution. Would anyone be able to explain this? Thanks!
Rankine Hugoniot condition for non-scalar initial data
partial differential equationswave equation
Related Solutions
It’s true! All the passages in the proof that you already know are reversible.
Edit:
Let $ [u] \stackrel{\text{df}}{=} {u_{+}}(p) - {u_{-}}(p) $ for all $ p \in \gamma_{0} = \{ (\xi(t),t) \mid t \in I \} $.
(R.H.S.): As $ \xi'(t) = \dfrac{[f(u)]}{[u]}(\xi(t),t) $ for all $ t \in I $, we have \begin{align} & [f(u)](\xi(t),t) - [u](\xi(t),t) \cdot \xi'(t) = 0 \\ \Longrightarrow \quad & 0 = \int_{I} \Big( [f(u)](\xi(t),t) - [u](\xi(t),t) \cdot \xi'(t) \Big) \cdot \phi(\xi(t),t) ~ \mathrm{d}{t}. \end{align}
Let $ v = (v_{1},v_{2}) = (1,- \xi'(t)) $. We can then write the previous formula as follows: \begin{align} 0 & = \int_{I} ([f(u)] v_{1} + [u] v_{2}) \phi ~ \mathrm{d}{s} \\ & = \int_{\gamma_{0} \cap \text{supp}(\phi)} \Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~ \mathrm{d}{s} - \int_{\gamma_{0} \cap \text{supp}(\phi)} \Big( f(u_{-}) v_{1} + u_{-} v_{2} \Big) \phi ~ \mathrm{d}{s}. \end{align}
We have taken $ \phi \in {C_{c}^{1}}(\Omega) $, so $ \text{supp}(\phi) = \omega_{-} \cup (\gamma_{0} \cap \text{supp}(\phi)) \cup \omega_{+} $.
This is the crucial point: We are going to use the Divergence Theorem in the ‘non-standard’ way: \begin{align} \int_{\gamma_{0} \cap \text{supp}(\phi)} \Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~ \mathrm{d}{s} & = \int_{\partial \omega_{+}} \Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~\mathrm{d}{s} \\ & = \iint_{\omega_{+}} \left( u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x} \right) ~ \mathrm{d}{x} \mathrm{d}{t} + \iint_{\omega_{+}} \left( \frac{\partial u}{\partial t} + \frac{\partial f(u)}{\partial x} \right) \phi ~ \mathrm{d}{x} \mathrm{d}{t}. \end{align}
Note that the last integral is $ 0 $ because our solution is classical outside the shock. We now have the following equations: $$ \int_{\gamma_{0} \cap \text{supp}(\phi)} \Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~ \mathrm{d}{s} = \iint_{\omega_{+}} \left( u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x} \right) ~ \mathrm{d}{x} \mathrm{d}{t}, $$ and, similarly, $$ - \int_{\gamma_{0} \cap \text{supp}(\phi)} \Big( f(u_{-}) v_{1} + u_{-} v_{2} \Big) \phi ~ \mathrm{d}{s} = \iint_{\omega_{-}} \left( u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x} \right) ~ \mathrm{d}{x} \mathrm{d}{t}. $$ Summing up term by term: \begin{align} 0 & = \int_{I} \Big( [f(u)] v_{1} + [u] v_{2} \Big) \phi ~ \mathrm{d}{s} \\ & = \iint_{\omega_{+} \cup \omega_{-}} \left( u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x} \right) ~ \mathrm{d}{x} \mathrm{d}{t} \\ & = \int_{\text{supp}(\phi)} \left( u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x} \right) ~ \mathrm{d}{x} \mathrm{d}{t}. \end{align} This is exactly what we were supposed to prove.
Hope it helps!
If needed, here is a sketch of the characteristic curves in $x$-$t$ plane, which may help to see better what happens:
The solution obtained by the method of characteristics up to the breaking time satisfies $u = \phi (x-ut)$ for $t<1$, i.e. $$ u(x,t) = \left\lbrace\begin{aligned} &1 & &\text{if}\quad x \leq t \\ &\tfrac{1-x}{1-t} & &\text{if}\quad t \leq x\leq 1\\ &0 & &\text{if}\quad x\geq 1 \end{aligned}\right. $$ At the breaking time, the shock speed $s$ is given by the Rankine-Hugoniot condition $s = \tfrac12 (1 + 0)$. Therefore, after the breaking time $t\geq 1$, $$ u(x,t) = \left\lbrace\begin{aligned} &1 & &\text{if}\quad x < 1 +\tfrac12 (t-1) \\ &0 & &\text{if}\quad x > 1 +\tfrac12 (t-1) \end{aligned}\right. $$
To your questions:
- The Rankine-Hugoniot condition $[\![\frac{1}{2}u^2]\!] = s [\![u]\!]$ is a condition satisfied by discontinuous weak solutions of the Burgers' equation. A weak solution which satisfies the Rankine-Hugoniot condition is not necessarily a solution to the proposed initial-value problem $u(x,0) = \phi(x)$. To be a solution, it must be in agreement with the characteristics equation until they cross, i.e. the fact that $u$ is constant along the curves $u = \phi(x - ut)$. Moreover, since weak solutions are not unique, it must satisfy an additional entropy condition. In the case of Burgers' equation, the Lax entropy condition amounts to the following statement: "if characteristics cross, then a shock-wave arises; but if they separate, a rarefaction wave occurs".
- The location of a discontinuity is deduced from the characteristics equation, whereas its speed is deduced from the Rankine-Hugoniot condition. In the present case, a shock occurs at the time $t=1$ (characteristics intersect). Its speed deduced from the Rankine-Hugoniot condition is $s=\frac{1}{2}$.
Best Answer
In the case of systems of conservation laws $$ \boldsymbol{u}_t + \boldsymbol{f}(\boldsymbol{u})_x = \boldsymbol{0} $$ where bold characters are vectors (it works also in the scalar case), the Rankine-Hugoniot condition (RH) for shock waves of speed $s$ is obtained by writing the scalar RH condition for each entry, i.e. $$ [\! [ \boldsymbol{f}(\boldsymbol{u}) ]\!] = s [\! [ \boldsymbol{u} ]\!] . $$ If $\boldsymbol{f}(\boldsymbol{u}) = \boldsymbol{A}\boldsymbol{u}+\boldsymbol{b}$ is linear, we end up with the eigenvalue problem $$ \boldsymbol{A} [\! [ \boldsymbol{u} ]\!] = s [\! [ \boldsymbol{u} ]\!] . $$ (The other query was answered directly in the comments of OP's linked post)